映射无形HList的类型

Question

我一直在尝试从scala的shapeless包中映射HList的类型，而无法访问它们的值。

以下成功映射了HList的值

import shapeless._
import shapeless.Poly._
import ops.hlist.Mapper
import ops.hlist.Mapper._

trait Person {
  type Value
  val v : Value
}

case class StringPerson extends Person {
  type Value = String
  val v = "I like strings"
}

case class IntPerson extends Person {
  type Value = Int 
  val v = 42
}

object what_is_going_on {

  object test_value_op {
    val stringPerson = StringPerson()
    val intPerson = IntPerson()

    trait lpvfun extends Poly1 {
      implicit def default[A <: Person] = at[A](_.v)
    } 

    object vfun extends lpvfun {}

    // Use these to generate compiler errors if the mapped type is not what we'd expect:

    type TestListType = StringPerson :: IntPerson :: HNil
    type TestListExpectedMappedType = String :: Int :: HNil

    // Input:
    val testList : TestListType = stringPerson :: intPerson :: HNil

    // Output:
    val mappedList : TestListExpectedMappedType = testList map vfun

    // Get the actual mapped type 
    type TestListActualMappedType = mappedList.type

    // This compiles......
    val mappedList1 : TestListActualMappedType = mappedList

    // .... but weirdly this line doesn't. That isn't the point of this question, but I'd be very grateful for an answer.
    //implicitly[TestListActualMappedType =:= TestListExpectedMappedType]
  }

}

凉！ 除了由于某种原因不能implicitly[A =:= B]使用implicitly[A =:= B] ，还已经映射了HList的值，因此也映射了它们的类型。

现在，假设我们没有HList值，但是我们知道它的类型。 我们如何映射其类型？

我根据此处 map的定义尝试了以下方法：

object test_type_op { 
  type TestListType = StringPerson :: IntPerson :: HNil
  type TestListExpectedMappedType = String :: Int :: HNil

  // Attempt 1 does not work, compiler cannot prove =:=
  type MappedType = Mapper[vfun.type, TestListType]#Out
  implicitly[MappedType =:= TestListExpectedMappedType]

  // Attempt 2 does not work, compiler cannot prove =:=
  class GetMapper {
    implicit val mapper : Mapper[vfun.type, TestListType]
    implicitly[mapper.Out =:= TestListExpectedMappedType]
  }

}

如何在不访问其值的情况下获取映射的HList的类型？ 有没有一种调试方法，为什么编译器无法证明某些内容？ 谢谢您的阅读。

Answer 1

在TestListActualMappedType的情况下，您获得了mappedList的单例类型，这与推断的mappedList类型mappedList 。 您可以看到完全相同的问题，而无需涉及Shapeless：

scala> val x = "foo"
x: String = foo

scala> implicitly[x.type =:= String]
<console>:13: error: Cannot prove that x.type =:= String.
       implicitly[x.type =:= String]
                 ^

您可以要求提供证据，证明x.type是String的子类型，或者可以使用shapeless.test.typed ，在您的情况下看起来像这样：

import shapeless._, ops.hlist.Mapper

trait Person {
  type Value
  val v : Value
}

case class StringPerson() extends Person {
  type Value = String
  val v = "I like strings"
}

case class IntPerson() extends Person {
  type Value = Int 
  val v = 42
}

trait lpvfun extends Poly1 {
  implicit def default[A <: Person] = at[A](_.v)
} 

object vfun extends lpvfun {}

val stringPerson = StringPerson()
val intPerson = IntPerson()

val testList = stringPerson :: intPerson :: HNil
val mappedList = testList map vfun

shapeless.test.typed[String :: Int :: HNil](mappedList)

不过，这与真正地指定类型相比并没有多大好处。

您可以要求提供证据，证明诸如Mapper类的类型类的输出类型是您期望的特定输入类型的类型：

scala> val m = Mapper[vfun.type, StringPerson :: IntPerson :: HNil]
m: shapeless.ops.hlist.Mapper[vfun.type,shapeless.::[StringPerson,shapeless.::[IntPerson,shapeless.HNil]]]{type Out = shapeless.::[String,shapeless.::[Int,shapeless.HNil]]} = shapeless.ops.hlist$Mapper$$anon$5@6f3598cd

scala> implicitly[m.Out =:= (String :: Int :: HNil)]
res1: =:=[m.Out,shapeless.::[String,shapeless.::[Int,shapeless.HNil]]] = <function1>

这更可能有用，但同样取决于您要说服自己的确切方式。